Engineering Mechanics - Statics Episode 2 Part 3 ppsx

Engineering Mechanics - Statics Chapter 6
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
kN 10
3
N=
Given:
P
1
4kN=
P
2
0kN=
a 2m=
θ
15 deg=
Solution:
Take advantage of the symetry.
Initial Guesses:
F
BD
1kN= F
CD
1kN= F
AB
1kN=
F
CA
1kN= F
BC

1kN=
Given
Joint D
P
1
−
2
F
BD
sin 2
θ
()
− F
CD
sin 3
θ
()
− 0=
Joint B
P
2
− cos 2
θ
()
F
BC
− 0=
F
BD
F

AB
− P
2
sin 2
θ
()
− 0=
Joint C
F
CD
cos
θ
()
F
CA
cos
θ
()
− 0=
F
CD
F
CA
+
()
sin
θ
()
F
BC

+ 0=
F
BD
F
CD
F
AB
F
CA
F
BC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find F

BD
F
CD
, F
AB
, F
CA
, F
BC
,
()
=
F
FD
F
ED
F
GF
F
EG
F
FE
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
F
BD
F
CD
F
AB
F
CA
F
BC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
481
Problem 6-27
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Positvive means Tension,
Negative means Compression
F
BD
F
CD
F
AB
F
CA
F
BC
⎛
⎜
⎜

⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
4−
0
4−
0
0
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟

⎟
⎟
⎟
⎠
kN=
F
FD
F
ED
F
GF
F
EG
F
FE
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎠
4−
0
4−
0
0
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
kN=
Problem 6-28
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
kN 10
3
N=

Given:
P
1
2kN=
P
2
4kN=
a 2m=
θ
15 deg=
Solution:
Take advantage of the symmetry.
Initial Guesses:
F
BD
1kN= F
CD
1kN= F
AB
1kN=
F
CA
1kN= F
BC
1kN=
Given
Joint D
P
1
−

2
F
BD
sin 2
θ
()
− F
CD
sin 3
θ
()
− 0=
Joint B
P
2
− cos 2
θ
()
F
BC
− 0=
F
BD
F
AB
− P
2
sin 2
θ
()

− 0=
Joint C
F
CD
cos
θ
()
F
CA
cos
θ
()
− 0=
F
CD
F
CA
+
()
sin
θ
()
F
BC
+ 0=
482
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6

F
BD
F
CD
F
AB
F
CA
F
BC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find F
BD

F
CD
, F
AB
, F
CA
, F
BC
,
()
=
F
FD
F
ED
F
GF
F
EG
F
FE
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
F
BD
F
CD
F
AB
F
CA
F
BC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎠
=
Positvive means Tension,
Negative means Compression
F
BD
F
CD
F
AB
F
CA
F
BC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
11.46−
6.69
13.46−
6.69
3.46−
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
kN=
F
FD

F
ED
F
GF
F
EG
F
FE
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
11.46−
6.69
13.46−
6.69

3.46−
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
kN=
Problem 6-29
Determine the force in each member of the truss and state if the members are in tension or
compression.
Units Used:
kip 10
3
lb=
Given:
F
1
2 kip=
F
2
1.5 kip=

F
3
3 kip=
F
4
3 kip=
a 4ft=
b 10 ft=
Solution:
θ
atan
a
b
⎛
⎜
⎝
⎞
⎟
⎠
=
Initial Guesses
483
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
F
AB
1lb= F
BC

1lb= F
CD
1lb
=
F
DE
1lb
=
F
AI
1lb= F
BI
1lb= F
CI
1lb= F
CG
1lb=
F
CF
1lb= F
DF
1lb= F
EF
1lb= F
HI
1lb=
F
GI
1lb= F
GH

1lb= F
FG
1lb=
Given
Joint A
F
AI
cos
θ
()
F
AB
+ 0=
Joint B
F
BC
F
AB
− 0=
F
BI
0=
Joint C
F
CD
F
BC
− F
CF
F

CI
−
()
cos
θ
()
+ 0=
F
CG
F
CF
F
CI
+
()
sin
θ
()
+ 0=
Joint D
F
DE
F
CD
− 0=
F
DF
0=
Joint I
F

2
F
GI
F
CI
+ F
AI
−
()
cos
θ
()
+ 0=
F
HI
F
BI
− F
GI
F
AI
− F
CI
−
()
sin
θ
()
+ 0=
Joint H

F
GH
cos
θ
()
F
1
+ 0=
F
GH
− sin
θ
()
F
HI
− 0=
Joint G
F
FG
F
GH
− F
GI
−
()
cos
θ
()
0=
F

3
− F
CG
− F
GH
F
FG
− F
GI
−
()
sin
θ
()
+ 0=
Joint F
F
EF
F
FG
− F
CF
−
()
cos
θ
()
0=
F
FG

F
CF
− F
EF
−
()
sin
θ
()
F
4
− F
DF
− 0=
484
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
F
AB
F
BC
F
CD
F
DE
F
AI
F

BI
F
CI
F
CG
F
CF
F
DF
F
EF
F
HI
F
GI
F
GH
F
FG
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find F
AB
F
BC
, F
CD
, F
DE
, F
AI
, F
BI
, F
CI
, F
CG
, F
CF
, F

DF
, F
EF
, F
HI
, F
GI
, F
GH
,
F
,
(
=
F
AB
F
BC
F
CD
F
DE
F
AI
⎛
⎜
⎜
⎜
⎜
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
3.75
3.75
7.75
7.75
4.04−
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟

⎠
kip=
F
BI
F
CI
F
CG
F
CF
F
DF
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠

0
0.27
1.4
4.04−
0
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
kip=
F
EF
F
HI
F
GI
F
GH
F
FG

⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
12.12−
0.8
5.92−
2.15−
8.08−
⎛
⎜
⎜
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎟
⎟
⎠
kip=
Positive means Tension, Negative means Compression
Problem 6-30
The Howe bridge truss is subjected to the loading shown. Determine the force in members DE,
EH, and HG, and state if the members are in tension or compression.
Units Used:
kN 10
3
N=
485
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Given:
F
1
30 kN=
F
2
20 kN=
F
3

20 kN=
F
4
40 kN=
a 4m=
b 4m=
Solution:
F
2
− aF
3
2a()− F
4
3a()− G
y
4a()+ 0=
G
y
F
2
2F
3
+ 3F
4
+
4
=
G
y
45kN=

Guesses F
DE
1kN= F
EH
1kN= F
HG
1kN=
Given
F
DE
− F
HG
− 0= G
y
F
4
− F
EH
− 0=
F
DE
bG
y
a+ 0=
F
DE
F
EH
F
HG

⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find F
DE
F
EH
, F
HG
,
()
=
F
DE
F
EH
F
HG
⎛
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎠
45−
5
45
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
Problem 6-31
The Pratt bridge truss is subjected to the loading shown. Determine the force in members LD,
LK, CD, and KD, and state if the members are in tension or compression.
Units Used:
kN 10
3
N=
486
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics Chapter 6
Given:
F
1
50 kN=
F
2
50 kN=
F
3
50 kN=
a 4m=
b 3m=
Solution:
A
x
0=
A
y
3F
3
4F
2
+ 5F
1
+
6
=
Guesses
F

LD
1kN= F
LK
1kN=
F
CD
1kN= F
KD
1kN=
Given
F
2
bF
1
2b()+ A
y
3b()− F
LK
a− 0=
F
CD
aF
1
b+ A
y
2b()− 0=
A
y
F
1

− F
2
−
a
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
LD
− 0=
F
3
− F
KD
− 0=
F
LD
F
LK
F
CD
F

KD
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
Find F
LD
F
LK
, F
CD
, F
KD
,
()
=
F
LD
F
LK

F
CD
F
KD
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
0
112.5−
112.5
50−
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
487
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Problem 6-32
The Pratt bridge truss is subjected to the loading shown. Determine the force in members JI,
JE, and DE, and state if the members are in tension or compression.
Units Used:
kN 10
3
N=
Given:
F
1
50 kN=
F
2
50 kN=
F
3
50 kN=
a 4m=

b 3m=
Solution:
Initial Guesses
G
y
1kN= F
JI
1kN=
F
JE
1kN= F
DE
1kN=
Given
Entire Truss
F
1
− bF
2
2b()− F
3
3b()− G
y
6b()+ 0=
Section
F
DE
− F
JI
− 0= F

JE
G
y
+ 0=
G
y
2b()F
DE
a− 0=
488
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
G
y
F
JI
F
JE
F
DE
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎟
⎟
⎠
Find G
y
F
JI
, F
JE
, F
DE
,
()
= G
y
50kN=
F
JI
F
JE
F
DE
⎛
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎠
75−
50−
75
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Positive means Tension,
Negative means Compression
Problem 6-33
The roof truss supports the vertical loading shown. Determine the force in members BC, CK,
and KJ and state if these members are in tension or compression.
Units Used:
kN 10
3
N=
Given:
F
1
4kN=

F
2
8kN=
a 2m=
b 3m=
489
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Solution:
Initial Guesses
A
x
1kN= A
y
1kN=
F
BC
1kN= F
CK
1kN=
F
KJ
1kN=
Given
A
x
0=
F

2
3a()F
1
4a()+ A
y
6a()− 0=
F
KJ
2b
3
⎛
⎜
⎝
⎞
⎟
⎠
A
x
2b
3
⎛
⎜
⎝
⎞
⎟
⎠
+ A
y
2a()− 0=
F

KJ
A
x
+
3a
b
2
9a
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
BC
+ 0=
F
CK
A
y
+
b
b
2
9a
2
+

⎛
⎜
⎝
⎞
⎟
⎠
F
BC
+ 0=
A
x
A
y
F
KJ
F
CK
F
BC
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟

⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
y
, F
KJ
, F
CK
, F
BC
,
()
=
A
x
A
y
F
KJ
F
CK
F
BC

⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
0
6.667
13.333
0
14.907−
⎛
⎜
⎜
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
Problem 6-34
Determine the force in members CD, CJ, KJ, and DJ of the truss which serves to support the
deck of a bridge. State if these members are in tension or compression.
Units Used:
kip 10
3
lb=
Given:
F
1
4000 lb=
F
2
8000 lb=
490
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
F

3
5000 lb=
a 9ft=
b 12 ft=
Solution
F
DJ
1 kip=
Initial Guesses:
A
y
1 kip= F
CD
1 kip=
F
CJ
1 kip= F
KJ
1 kip=
Given
F
3
aF
2
4a()+ F
1
5a()+ A
y
6a()− 0=
A

y
− 2a()F
1
a+ F
KJ
b+ 0=
F
CD
F
KJ
+
a
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
CJ
+ 0=
A
y
F
1

− F
2
−
b
a
2
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
F
CJ
− 0=
F
DJ
− 0=
A
y
F
KJ
F
CJ
F
DJ
F

CD
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
y
F
KJ
, F
CJ
, F
DJ
, F
CD
,

()
=
A
y
F
KJ
F
CJ
F
DJ
F
CD
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠

9.5
11.25
3.125−
0
9.375−
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
kip=
Positive (T)
Negative (C)
491
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Determine the force in members EI and JI of the truss which serves to support the deck of a
bridge. State if these members are in tension or compression.
Units Used:

kip 10
3
lb=
Given:
F
1
4000 lb=
F
2
8000 lb=
F
3
5000 lb=
a 9ft=
b 12 ft=
Solution:
Initial Guesses:
G
y
1 kip= F
EI
1 kip= F
JI
1 kip=
Given
F
1
− aF
2
2a− F

3
5a− G
y
6a+ 0=
G
y
2aF
3
a− F
JI
b− 0=
F
EI
F
3
− G
y
+ 0=
G
y
F
JI
F
EI
⎛
⎜
⎜
⎜
⎝
⎞

⎟
⎟
⎟
⎠
Find G
y
F
JI
, F
EI
,
()
=
G
y
F
JI
F
EI
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
7.5

7.5
2.5−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kip=
Positive (T)
Negative (C)
Problem 6-36
Determine the force in members BE, EF, and CB, and state if the members are in tension or
compression.
Units Used:
kN 10
3
N=
492
Problem 6-35
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Given:
F
1
5kN= F

4
10 kN=
F
2
10 kN= a 4m=
F
3
5kN= b 4m=
Solution:
θ
atan
a
b
⎛
⎜
⎝
⎞
⎟
⎠
=
Inital Guesses
F
CB
1kN= F
BE
1kN= F
EF
1kN=
Given
F

1
F
2
+ F
BE
cos
θ
()
− 0=
F
CB
− F
EF
− F
BE
sin
θ
()
− F
3
− 0=
F
1
− aF
CB
b+ 0=
F
CB
F
BE

F
EF
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find F
CB
F
BE
, F
EF
,
()
=
F
CB
F
BE
F
EF
⎛
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎠
5
21.2
25−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
Problem 6-37
Determine the force in members BF, BG, and AB, and state if the members are in tension or
compression.
Units Used:
kN 10
3
N=
Given:
F

1
5kN= F
4
10 kN=
493
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
F
2
10 kN= a 4m=
F
3
5kN= b 4m=
Solution:
θ
atan
a
b
⎛
⎜
⎝
⎞
⎟
⎠
=
Inital Guesses
F
AB

1kN= F
BG
1kN= F
BF
1kN=
Given
F
1
F
2
+ F
4
+ F
BG
cos
θ
()
+ 0=
F
1
− 3aF
2
2a− F
4
a− F
AB
b+ 0=
F
BF
− 0=

F
AB
F
BG
F
BF
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find F
AB
F
BG
, F
BF
,
()
=
F
AB
F
BG
F

BF
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
45
35.4−
0
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
Problem 6-38
Determine the force developed in members GB and GF of the bridge truss and state if these
members are in tension or compression.
Given:
F

1
600 lb=
494
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
F
2
800 lb=
a 10 ft=
b 10 ft=
c 4ft=
Solution:
Initial Guesses
A
x
1lb= A
y
1lb=
F
GB
1lb= F
GF
1lb=
Given
F
2
bF
1

b 2c+()+ A
y
2 bc+()− 0=
A
x
0=
A
y
F
GB
− 0=
A
y
− bF
GF
a− 0=
A
x
A
y
F
GB
F
GF
⎛
⎜
⎜
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
Find A
x
A
y
, F
GB
, F
GF
,
()
=
A
x
A
y
F
GB
F
GF
⎛
⎜
⎜

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
0
671.429
671.429
671.429−
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
lb=
Positive (T)
Negative (C)

Problem 6-39
Determine the force members BC, FC, and FE, and state if the members are in tension or
compression.
Units Used:
kN 10
3
N=
495
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Given:
F
1
6kN=
F
2
6kN=
a 3m=
b 3m=
Solution:
θ
atan
a
b
⎛
⎜
⎝
⎞

⎟
⎠
=
Initial Guesses
D
y
1kN= F
BC
1kN=
F
FC
1kN= F
FE
1kN=
Given
F
1
− bF
2
2b()− D
y
3b()+ 0=
D
y
bF
FE
cos
θ
()
a− 0=

F
FC
− F
BC
F
FE
+
()
cos
θ
()
− 0=
F
2
− D
y
+ F
FE
F
BC
+
()
sin
θ
()
+ 0=
D
y
F
BC

F
FC
F
FE
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
Find D
y
F
BC
, F
FC
, F
FE
,
()
=
D

y
F
BC
F
FC
F
FE
⎛
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
6
8.49−
0
8.49
⎛
⎜
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
496
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Determine the force in members IC and CG of the truss and state if these members are in tension
or compression. Also, indicate all zero-force members.
Units Used:
kN 10
3
N=
Given:
F
1
6kN=
F
2
6kN=
a 1.5 m=
b 2m=

Solution:
By inspection of joints B, D, H and I.
AB, BC, CD, DE, HI, and GI are all zero-force members.
Guesses A
y
1kN= F
IC
1kN= F
CG
1kN= F
CJ
1kN=
Given A
y
− 4a()F
1
2a()+ F
2
a+ 0=
A
y
− 2a()
b
a
2
b
2
+
F
IC

a−
a
a
2
b
2
+
F
IC
b− 0=
a−
a
2
b
2
+
F
IC
a
a
2
b
2
+
F
CJ
+ 0=
b−
a
2

b
2
+
F
IC
b
a
2
b
2
+
F
CJ
− F
CG
− 0=
A
y
F
IC
F
CG
F
CJ
⎛
⎜
⎜
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
Find A
y
F
IC
, F
CG
, F
CJ
,
()
=
A
y
F
IC
F
CG
F
CJ
⎛
⎜
⎜

⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
4.5
5.625−
9
5.625−
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)

Problem 6-41
Determine the force in members JE and GF of the truss and state if these members are in
tension or compression. Also, indicate all zero-force members.
497
Problem 6-40
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Units Used:
kN 10
3
N=
Given:
F
1
6kN=
F
2
6kN=
a 1.5 m=
b 2m=
Solution:
By inspection of joints B, D, H and I.
AB, BC, CD, DE, HI, and GI are all zero-force members.
Guesses E
y
1kN= F
JE
1kN= F

GF
1kN=
Given F
1
− 2a()F
2
3a()− E
y
4a()+ 0=
E
y
b
a
2
b
2
+
F
JE
+ 0=
a−
a
2
b
2
+
F
JE
F
GF

− 0=
E
y
F
JE
F
GF
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
Find E
y
F
JE
, F
GF
,
()
=
E
y
F
JE

F
GF
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
7.5
9.375−
5.625
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
Problem 6-42
Determine the force in members BC, HC, and HG. After the truss is sectioned use a single
equation of equilibrium for the calculation of each force. State if these members are in tension
or compression.

Units Used:
kN 10
3
N=
498
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Given:
F
1
2kN= F
4
5kN= a 5m=
F
2
4kN= F
5
3kN= b 2m=
F
3
4kN= c 3m=
Solution:
Guesses
A
x
1kN= A
y
1kN=

F
BC
1kN= F
HC
1kN=
F
HG
1kN= d 1m=
Given
c
ad+
b
a
= A
x
− 0=
F
1
A
y
−
()
4a()F
2
3a()+ F
3
2a()+ F
4
a()+ 0=
F

1
A
y
−
()
a() A
x
c()+ F
BC
c()− 0=
F
1
A
y
−
()
2a()F
2
a()+
a
a
2
b
2
+
F
HG
c()+
b
a

2
b
2
+
F
HG
a()+ 0=
A
y
F
1
−
()
d() F
2
ad+()−
c
a
2
c
2
+
F
HC
ad+()+
a
a
2
c
2

+
F
HC
c()+ 0=
A
y
A
x
F
BC
F
HC
F
HG
d
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎟
⎠
Find A
y
A
x
, F
BC
, F
HC
, F
HG
, d,
()
=
A
x
A
y
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠

0
8.25
⎛
⎜
⎝
⎞
⎟
⎠
kN=
F
BC
F
HC
F
HG
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
10.417−
2.235
9.155
⎛
⎜

⎜
⎝
⎞
⎟
⎟
⎠
kN=
d 2.5 m=
Positive (T)
Negative (C)
Problem 6-43
Determine the force in members CD, CF, and CG and state if these members are in tension or
499
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
compression.
Units Used:
kN 10
3
N=
Given:
F
1
2kN= F
4
5kN= a 5m=
F
2

4kN= F
5
3kN= b 2m=
F
3
4kN= c 3m=
Solution:
Guesses
E
y
1kN= F
CD
1kN=
F
CF
1kN= F
CG
1kN=
F
FG
1kN= F
GH
1kN=
Given
F
2
− a() F
3
2a()− F
4

3a()− E
y
F
5
−
()
4a()+ 0=
F
CD
c() E
y
F
5
−
()
a()+ 0=
F
4
− a() F
5
E
y
−
()
2a()−
a
a
2
b
2

+
F
FG
bc+()− 0=
a
a
2
b
2
+
F
FG
a
a
2
b
2
+
F
GH
− 0=
b
a
2
b
2
+
F
FG
F

GH
+
()
F
CG
+ 0=
F
5
E
y
−
()
ac b−()
b
F
4
a
ac b−()
b
+
⎡
⎢
⎣
⎤
⎥
⎦
+
c
a
2

c
2
+
F
CF
2 a
ac b−()
b
+
⎡
⎢
⎣
⎤
⎥
⎦
− 0=
500
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
E
y
F
CD
F
CF
F
CG
F

FG
F
GH
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find E
y
F
CD
, F
CF
, F

CG
, F
FG
, F
GH
,
()
=
E
y
F
CD
F
CF
F
CG
F
FG
F
GH
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
9.75
11.25−
3.207
6.8−
9.155
9.155
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟

⎟
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
Problem 6-44
Determine the force in members OE, LE, and LK of the Baltimore truss and state if the
members are in tension or compression.
Units Used:
kN 10
3
N=
Given:
F
1
2kN= a 2m=
F
2
2kN= b 2m=
F
3
5kN=
F
4
3kN=
Solution:
A
x

0kN=
Initial Guesses
A
y
1kN= F
OE
1kN=
F
DE
1kN= F
LK
1kN=
F
LE
1kN=
Given
F
LE
0=
F
4
3b()F
3
4b()+ F
2
5b()+ F
1
6b()+ A
y
8b()− 0=

F
LK
F
DE
+ F
OE
b
a
2
b
2
+
+ 0=
501
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
A
y
F
1
− F
2
− F
OE
a
a
2
b

2
+
− 0=
F
LK
− 2a()F
2
b()+ F
1
2b()+ A
y
4b()− 0=
A
y
F
OE
F
DE
F
LK
F
LE
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜

⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Find A
y
F
OE
, F
DE
, F
LK
, F
LE
,
()
=
A
y
F
DE
⎛
⎜
⎜

⎝
⎞
⎟
⎟
⎠
6.375
7.375
⎛
⎜
⎝
⎞
⎟
⎠
kN=
F
OE
F
LE
F
LK
⎛
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠

3.36
0
9.75−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kN=
Positive (T)
Negative (C)
Problem 6-45
Determine the force in member GJ of the truss and state if this member is in tension or
compression.
Units Used:
kip 10
3
lb=
Given:
F
1
1000 lb=
F
2
1000 lb=
F
3

1000 lb=
F
4
1000 lb=
a 10 ft=
θ
30 deg=
502
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Guess E
y
1lb= F
GJ
1lb=
Given
F
2
− a() F
3
2a()− F
4
3a()− E
y
4a()+ 0=
F
4
− a() E

y
2a()+ F
GJ
sin
θ
()
2a()+ 0=
E
y
F
GJ
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
Find E
y
F
GJ
,
()
=
E
y
F
GJ

⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
1.5
2−
⎛
⎜
⎝
⎞
⎟
⎠
kip=
Positive (T)
Negative (C)
Problem 6-46
Determine the force in member GC of the truss and state if this member is in tension or
compression.
Units Used:
kip 10
3
lb=
Given:
F
1
1000 lb=

F
2
1000 lb=
F
3
1000 lb=
F
4
1000 lb=
a 10 ft=
θ
30 deg=
Solution:
Guess E
y
1lb= F
GJ
1lb=
F
HG
1lb= F
GC
1lb=
Given
F
2
− a() F
3
2a()− F
4

3a()− E
y
4a()+ 0=
F
4
− a() E
y
2a()+ F
GJ
sin
θ
()
2a()+ 0=
503
=
Solution:
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
F
HG
− cos
θ
()
F
GJ
cos
θ
()

+ 0=
F
3
− F
GC
− F
HG
F
GJ
+
()
sin
θ
()
− 0=
E
y
F
GJ
F
GC
F
HG
⎛
⎜
⎜
⎜
⎜
⎜
⎝

⎞
⎟
⎟
⎟
⎟
⎟
⎠
Find E
y
F
GJ
, F
GC
, F
HG
,
()
=
E
y
F
GJ
F
GC
F
HG
⎛
⎜
⎜
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
1.5
2−
1
2−
⎛
⎜
⎜
⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎠
kip=
Positive (T)
Negative (C)
Problem 6-47

Determine the force in members KJ, JN, and CD, and state if the members are in tension or
compression. Also indicate all zero-force members.
Units Used:
kip 10
3
lb=
Given:
F 3 kip=
a 20 ft=
b 30 ft=
c 20 ft=
Solution:
A
x
0=
θ
atan
2c
2ab+
⎛
⎜
⎝
⎞
⎟
⎠
=
φ
atan
2c
b

⎛
⎜
⎝
⎞
⎟
⎠
=
Initial Guesses:
A
y
1lb= F
CD
1lb=
F
KJ
1lb= F
JN
1lb=
504
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 6
Fa
b
2
+
⎛
⎜
⎝

⎞
⎟
⎠
A
y
2ab+()− 0=
F
CD
cA
y
a
b
2
+
⎛
⎜
⎝
⎞
⎟
⎠
− 0=
F
CD
F
JN
cos
φ
()
+ F
KJ

cos
θ
()
+ 0=
A
y
F
JN
sin
φ
()
+ F
KJ
sin
θ
()
+ 0=
A
y
F
CD
F
JN
F
KJ
⎛
⎜
⎜
⎜
⎜

⎜
⎝
⎞
⎟
⎟
⎟
⎟
⎟
⎠
Find A
y
F
CD
, F
JN
, F
KJ
,
()
= A
y
1.5 kip=
F
CD
F
JN
F
KJ
⎛
⎜

⎜
⎜
⎝
⎞
⎟
⎟
⎟
⎠
2.625
0
3.023−
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
kip=
Positive (T), Negative (C)
Problem 6-48
Determine the force in members BG, HG, and BC of the truss and state if the members are in
tension or compression.
Units Used:
kN 10
3
N=
Given:
F

1
6kN=
F
2
7kN=
F
3
4kN=
a 3m=
b 3m=
c 4.5 m=
505
Given
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.